Question 1158809



{{{y = a*b^x}}}......use the point ({{{-1}}}, {{{1/2}}} )

{{{1/2 = a*b^-1}}}

{{{1/2 = a/b}}}

{{{b=2a}}}..........eq.1


{{{32 = a*(2a)^2}}}.....using the point   ({{{2}}},{{{32}}})

{{{32= a*4a^2}}}

{{{32=4a^3}}}

{{{8=a^3}}}

{{{a=root(3,8)}}}

{{{a=root(3,2^3)}}}

{{{a= 2}}}.........eq.2


substitute in eq.1 


{{{b=2a}}}..........eq.1
{{{b=2*2}}}
{{{b=4}}}


your equation is:

{{{y =  2*4^x}}}


{{{drawing( 600, 600, -10, 10, -10, 45,
circle(-1,1/2,.12),locate(-1,1/2,p(-1,1/2)),circle(2,32,.12),locate(2,32,p(2,32)),
graph( 600, 600, -10, 10, -10, 45, 2*4^x)) }}}